1. Apr 18, 2006

mathboy20

Hi

I'm told that the the power series:

$$\sum_{n=0} ^ \infty (2n+1) z^n$$ has the radius of convergens

R = 1.

Proof:

Using the Definition of convergens for power series:

$$\frac{(2n+1)}{(2n+1)+1} = \frac{(2n+1)}{(2n+3)}$$

$$limit _{n \rightarrow \infty} \frac{(2n+1)}{(2n+3)} = 1$$

Therefore the radius of convergens is R = 1. Right ?

Second question: The Power series above suposedly diverges on every point on the circle of convergens. How do I show that?

I know that according to the definition of divergens of the power series:

$$\sum_{n = 0} ^{\infty} a_n z^n$$

that $$a_n \rightarrow \infty$$ for $$n \rightarrow \infty$$

Do I the use this fact here to show that a_n diverges ??

Best Regards
Mathboy20

Last edited: Apr 18, 2006
2. Apr 18, 2006

Jameson

I'm not following this step. I would use the ratio test for finding the radius of convergence.

3. Apr 20, 2006

kam.epi

I think this would be the proper way to evaluate the series:

$$\sum_{n=0} ^ \infty (2n+1) z^n$$

so using the ratio test,

$$limit _{n \rightarrow \infty} \frac{(2(n+1)+1) z^(n+1)}{(2n+1) z^n}$$

$$limit _{n \rightarrow \infty} \frac{(2n+3)}{(2n+1)}|z|$$

Therefore, $$|z|<1$$

implies that the interval of convergence is $$-1<z<1$$

Therefore, the radius of convergence must be 1.

4. Apr 23, 2006

mathboy20

Anyway if I then have show that the series diverges for all point on the circle of convergens.

Doesn't that mean that

$$\frac{2n+3}{2n+1}|z| \geq 1$$, where $$n \neq 0$$

if n = 1

then $$|z| \geq \frac{3}{5}$$

Am I on the right track here?

Best Regards
Mathboy20

Last edited: Apr 23, 2006
5. Apr 23, 2006

HallsofIvy

Staff Emeritus
n= 1 is not relevant. What is
$$lim_{n\rightarrow \infty}\frac{2n+3}{2n+1}$$?